Find the number of real solutions to the equation $2^x+3^x+4^x=5^x+6^x$.

Solution

1. Divide by $6^x$: $(2/6{)}^x+(3/6{)}^x+(4/6{)}^x=(5/6{)}^x+1$.
2. Let $f(x)=(1/3{)}^x+(1/2{)}^x+(2/3{)}^x-(5/6{)}^x-1$.
3. Since each term $(a/b{)}^x$ is strictly decreasing for $a<b$, $f(x)$ is strictly decreasing.
4. As $x\to -\infty$, $f(x)\to \infty$, and as $x\to \infty$, $f(x)\to -1$.
5. By the Intermediate Value Theorem, there is exactly one root.
6. Hence the answer is (B).